ADC - 2 channels but not same voltage reference - how to?

Hello,

It's a lead-acid battery charger project and I'd like to control the voltage applied to the batteries but not in the full charger's range (0V-32V) for accuracy reasons.

I measure the load's current via R 0,1ohm; no problem here (I have made a 10x gain ampli-op circuit to get a usable reading). Reference for this ADC channel is GND.

For the voltage range that will be controlled, the second ADC channel would return a value between 195 (25V) and 250 (32V) if I use the same voltage reference (using a 2 R divider connected between 0 and 32VDC). BUT, doing so, I use only around 20% of the ADC's resolution range.

I want to monitor a range of approx 7 volts (lets say from 25V to 32V) because this is the range I am really going to control the batteries's charging voltage. I would then (theoretically) read ADC values from (approx) 5 to 250 wich would be almost the full range (hope I'm clear enough...).

My idea is to use a 25V Zener.
<img src="http://home.citycable.ch/flotulopex/downloads/I-U-meas.jpg">

The problem is that both channels won't have the same voltage reference (GND).

How do I resolve this?

Hi Roger,
I don’t know how to separate the voltage reference grounds on PIC’s ADC channels. It would be interesting to find out if it is possible.

You could move the Zener to the “high side” (32VDC) and then measure between GND and the 25 Zener’s anode. This would subtract the 25 Volts. You would still need to “voltage divide” the ~7V, to fit the limits of the PIC ADC.

Both of these schemes cause a small error introduced by the tiny drop on the 0,1 Ohm shunt resistor.

You could reference both…. at the 0,1 Ohm shunt resistor / Battery junction.
Then, make your 10x gain amplifier-op circuit an inverter to get a positive voltage out, for the “current” channel. You would have to isolate the PIC’s power supply and have a plus/minus supply for the op-amp.

-Adam-

Well, I should have think about this solution already...

You must be right.

I'm gonna try inverting the Zenner's position.

Thanks.